Đáp án:
\(\begin{array}{l}
b)\\
{m_{N{a_2}S{O_4}}} = 14,2g\\
c)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 49g\\
d)\\
C{\% _{N{a_2}S{O_4}}} = 24,91\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
b)\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{8}{{40}} = 0,2mol\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{N{a_2}S{O_4}}} = 0,1 \times 142 = 14,2g\\
c)\\
{n_{{H_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{9,8 \times 100}}{{20}} = 49g\\
d)\\
{m_{{\rm{dd}}spu}} = 8 + 49 = 57g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{14,2}}{{57}} \times 1005 = 24,91\%
\end{array}\)
